【UOJ188】【UR #13】Sanrd

【题目链接】

• 点击打开链接

【思路要点】

• 令$f(x)$表示$x$次大的质因子，那么答案即为$\sum_{i=l}^{r}f(i)$，也就是说我们要求的实际上是$\sum_{i=1}^{N}f(i)$。

• 考虑从小到大枚举每一个质因子，最后第二个被枚举到的质因子即为$f(x)$。

• 类似$Min25$筛的过程，我们令$s(N,i)$表示上次选择的质因子为$prime_{i-1}$，剩余部分的乘积不超过$N$的数的$f(x)$之和。

• 有

  $$s(N,i)=\left\{\begin{array}{rcl}0 & & {prime_i>N}\\prime_{i-1}*(\sum_{j=1}^{N}[j\ is\ a\ prime]-(j-1))\\+\sum_{j=i}^{prime_j^2≤N}\sum_{k=1}^{prime_j^{k+1}≤N}(s(\lfloor\frac{N}{prime_j^k}\rfloor,j+1)+prime_j) & & {prime_i≤N}\end{array} \right.$$

• 那么$\sum_{i=1}^{N}f(i)=s(N,1)$。

• 时间复杂度$O(\frac{N^{\frac{3}{4}}}{LogN})$。

【代码】

#include<bits/stdc++.h>

using namespace std;
const int MAXN = 700005;
typedef unsigned long long ull;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
  x = 0; int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  x *= f;
}
template <typename T> void write(T x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
  write(x);
  puts("");
}
ull n, limit, val[MAXN], cnt[MAXN];
int m, tot, prime[MAXN], f[MAXN], home1[MAXN], home2[MAXN];
void init(int n) {
  tot = 0;
  for (int i = 2; i <= n; i++)
      f[i] = 0;
  for (int i = 2; i <= n; i++) {
      if (f[i] == 0) prime[++tot] = f[i] = i;
      for (int j = 1; j <= tot && prime[j] <= f[i]; j++) {
          int tmp = prime[j] * i;
          if (tmp > n) break;
          f[tmp] = prime[j];
      }
  }
}
ull s(ull x, int y) {
  if (x <= 1 || prime[y] > x) return 0;
  ull ans = 0;
  if (x <= limit) ans = cnt[home1[x]] - (y - 1);
  else ans = cnt[home2[n / x]] - (y - 1);
  ans *= prime[y - 1];
  for (int i = y; i <= tot && 1ull * prime[i] * prime[i] <= x; i++) {
      ull now = prime[i], nxt = now * now;
      for (int j = 1; nxt <= x; j++, now *= prime[i], nxt *= prime[i])
          ans += s(x / now, i + 1) + prime[i];
  }
  return ans;
}
ull solve(ull x) {
  n = x;
  limit = sqrt(n);
  init(limit);
  m = 0;
  for (ull i = 1, nxt; i <= n; i = nxt) {
      ull tmp = n / i;
      nxt = n / tmp + 1;
      val[++m] = tmp;
      cnt[m] = tmp - 1;
      if (tmp <= limit) home1[tmp] = m;
      else home2[i] = m; 
  }
  for (int i = 1; i <= tot; i++)
  for (int j = 1; 1ull * prime[i] * prime[i] <= val[j]; j++) {
      ull tmp = val[j] / prime[i];
      if (tmp <= limit) cnt[j] -= cnt[home1[tmp]] - (i - 1);
      else cnt[j] -= cnt[home2[n / tmp]] - (i - 1);
  }
  return s(n, 1);
}
int main() {
  ull l, r; read(l), read(r);
  writeln(solve(r) - solve(l - 1));
  return 0;
}